 * refreshed formula sheet
* fixed page layout somewhat
 \renewcommand{\lastedityear}{2019} \renewcommand{\lasteditmonth}{03} \renewcommand{\lasteditday}{26} \renewcommand{\lasteditday}{30} \atstartofexercises \fluidmechexercisestitle ... ... @@ -10,35 +10,30 @@ \mecafluexboxen \begin{boiboiboite} Reynolds Transport Theorem: \begin{IEEEeqnarray}{rCl} \timederivative{B_\sys} & = & \timederivative{} \iiint_\cv \rho b \diff \vol + \iint_\cs \rho b \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt}} \end{IEEEeqnarray} Mass conservation: \begin{IEEEeqnarray}{rCCCCCl} \timederivative{m_\sys} & = & 0 & = & \timederivative{} \iiint_\cv \rho \diff \vol & + & \iint_\cs \rho \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_mass}} \end{IEEEeqnarray} Change in linear momentum: \begin{IEEEeqnarray}{rCCCCCl} \timederivative{(m \vec V_{sys})} & = & \vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol & + & \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_linearmom}} \end{IEEEeqnarray} Change in angular momentum: \begin{equation} \timederivative{(\vec r_{\X m} \wedge m \vec V)_\sys} = \vec M_{\net, \X} = \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol + \iint_\cs \vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V \diff A \tag{\ref{eq_rtt_angularmom}} \end{equation} Mass balance through an arbitrary volume: \begin{IEEEeqnarray}{rCCCl} 0 & = & \timederivative{} \iiint_\cv \rho \diff \vol & + & \iint_\cs \rho \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_mass}} \end{IEEEeqnarray} Momentum balance through an arbitrary volume: \begin{IEEEeqnarray}{rCCCl} \vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol & + & \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_linearmom}} \end{IEEEeqnarray} Angular momentum balance through an arbitrary volume: \begin{IEEEeqnarray}{rCCCl} \vec M_{\net, \X} &=& \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol &+& \iint_\cs \vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V \diff A \ztag{\ref{eq_rtt_angularmom}} \end{IEEEeqnarray} \end{boiboiboite} \clearpage%handmade %%%% \subsubsection{Pipe bend} \wherefrom{\cczero \oc} \label{exo_pipe_bend} A pipe with diameter~\SI{30}{\milli\meter} has a bend with angle $\theta = \SI{130}{\degree}$, as shown in \cref{fig_pipe_bend}. Water enters and leaves the pipe with the same speed $V_1 = V_2 = \SI{1,5}{\metre\per\second}$. The velocity distribution at both inlet and outlet is uniform. \begin{figure} \begin{figure}[ht!] \begin{center} \includegraphics[width=7.5cm]{pipe_bend} \vspace{-0.5cm} ... ... @@ -53,9 +48,6 @@ Change in angular momentum: \item What would be the new force if all of the speeds were doubled? \end{enumerate} %%%% \subsubsection{Exhaust gas deflector} \label{exo_exhaust_gas_deflector} ... ... @@ -86,8 +78,7 @@ Change in angular momentum: \label{exo_pelton_turbine} A water turbine is modeled as the following system: a water jet exiting a stationary nozzle hits a blade which is mounted on a rotor (\cref{fig_water_turbine}). In the ideal case, viscous effects can be neglected, and the water jet is deflected entirely with a~\SI{180}{\degree} angle. \begin{figure} \begin{figure}[ht!] \begin{center} \includegraphics[width=12cm]{water_turbine_blade} \end{center} ... ... @@ -258,6 +249,7 @@ Change in angular momentum: %%%% \subsubsection{Moment on gas deflector} \label{exo_moment_gas_deflector} \wherefrom{non-examniable} We revisit the exhaust gas deflector of exercise \ref{exo_exhaust_gas_deflector} p.\pageref{exo_exhaust_gas_deflector}. \Cref{fig_deflector_sideview} below shows the deflector viewed from the side. The midpoint of the inlet is \SI{2}{\metre} above and \SI{5}{\metre} behind the wheel labeled~“\textbf{A}”, while the midpoint of the outlet is \SI{3,5}{\metre} above and \SI{1,5}{\metre} behind it. \begin{figure}[ht!] ... ... @@ -270,10 +262,11 @@ Change in angular momentum: What is the moment generated by the gas flow about the axis of the wheel labeled “\textbf{A}”? \clearpage%handmade %%%% \subsubsection{Helicopter tail moment} \label{exo_helicopter_tail_moment} \wherefrom{non-examinable} In a helicopter, the role of the tail is to counter exactly the moment exerted by the main rotor about the main rotor axis. This is usually done using a tail rotor which is rotating around a horizontal axis. ... ... @@ -297,7 +290,7 @@ Change in angular momentum: \item Propose and quantify a modification to the tail geometry or operating conditions that would allow the tail to produce no thrust (that is to say, zero force in the $x$-axis), while still generating the same moment. \end{enumerate} \textit{Remark: this system is commercialized by MD Helicopters as the \wed{NOTAR}{\textsc{notar}}. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this effect is explored in chapter~8 (\S\ref{ch_circulating_cylinder} p.\pageref{ch_circulating_cylinder}).} \textit{Remark: this system is commercialized by MD Helicopters as the \wed{NOTAR}{\textsc{notar}}. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this effect is explored in \chaptereleven (\S\ref{ch_circulating_cylinder} p.\pageref{ch_circulating_cylinder}).} %%%% ... ... @@ -407,7 +400,7 @@ Change in angular momentum: \tab $V_\text{center} = \num{1,2245} U$; $F_\net = \SI{+393}{\newton}$ (positive!) \item [\ref{exo_thrust_reverser}]% \tab $\dot m_\text{cold} = \SI{297}{\kilogram\per\second}$, $\dot m_\text{hot} = \SI{59,4}{\kilogram\per\second}$; \tab $F_\text{cold flow, normal, bench \& runway} = \SI{+74,25}{\kilo\newton}$, $F_\text{hot flow, normal \& reverse, bench \& runway} = \SI{+8,316}{\kilo\newton}$; \tab $F_\text{cold flow, normal, bench \& runway} = \SI{+74,25}{\kilo\newton}$,\\ $F_\text{hot flow, normal \& reverse, bench \& runway} = \SI{+8,316}{\kilo\newton}$; \tab $F_\text{cold flow, reverse, bench \& runway} = \SI{-24,35}{\kilo\newton}$ and $F_\text{hot flow, normal, bench \& runway} = \SI{+8,316}{\kilo\newton}$. \tab Adding the net pressure force due to the (lossless) flow acceleration upstream of the inlet, we obtain, on the bench: $F_\text{engine bench, normal} = \SI{-93,26}{\kilo\newton}$, $F_\text{engine bench, reverse} = \SI{+5,344}{\kilo\newton}$; and on the runway: $F_\text{engine runway, normal} = \SI{-92,5}{\kilo\newton}$ and $F_\text{engine bench, reverse} = \SI{+6,094}{\kilo\newton}$. ... ...